MO-ARML --- September, POWER OF A POINT

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1 M-ML --- Septembe, W INT owe of a oint is a set of thee-theoems-in-one about cicles and line segments. * = * 2 = * * = * XISS G S X H Z 3 6 H 7 T K. = 4 and X < X, find X.. ind HK.. ind TV. U V 2. THM: If,,, and ae on a cicle then fo any point in the cicle, * = *. ove.. aw and. Why does =?. Which two tiangles ae simila? ove it.. omplete the poof that * = *. 3. ften othe theoems must be used along with owe of a oint. = 8; = 0; = 7; = 60 o. etemine the aea of the cicle. HSM, 992, #27

2 W INT ---- WKSHT.. Z Y. 6 2 H W X K = 2; = 2; = 2; find Y = 4; YZ = 8; WX = 2; find W K is a ight angle. ind K. 2. = and = =. ind / 3. oint is exteio to cicle with adius. line though intesects the cicle at and. etemine * in tems of and. 3. epeat if is inteio to cicle. 4. Two cicles intesect at points and. is on line such that and Q ae tangents to the cicle. V: = Q. Q. oint is exteio to a cicle with as a tangent. line though intesects the cicle at points and. V: 2 = *. aw segments and. Why does =? 2. Which two tiangles ae simila? Why? 3. omplete the poof.. second line though intesects the cicle at points and. V the oollay: * = * [eing a coollay suggests that thee is a vey easy, shot poof!]

3 6. is pependicula to and equals the length of the iamete of cicle. ompute /. 7. and ae points on a cicle with cente. lies outside the cicle, on ay. Given = 24, = 28, and =, find. Hint: The solution to a ecent poblem is helpful hee. 8. quilateal tiangle is inscibed in a cicle. Q is on segment. is on line Q and on the cicle. V: = + Q 9. n equilateal tiangle is inscibed in a cicle. and ae midpoints of and espectively. is the point whee ay intesects the cicle. ompute /. 0. In the figue, is a quadilateal with ight angles at and. and ae on such that and ae pependicula to. If = 3, =, and = 7, compute.

4 NSWS with HINTS of solutions XISS. XG*XH = X*X Let x = X, then X = 4 x; 8*3 = x(4 x) x 2 4x + 24 = (x -2)(x 2) = 0. X = 2 o 2 Since X < X, X = = H * K, Let HK = x, then K = x + 6; 8*8 = 6 * (x + 6); x = 4/3. ZS * ZU = ZT * ZV; Let TV = x, then ZV = 7 + x. * 7 = 7 * (7 + x) x = 36/7 2. aw. ngle and angle each subtend the same chod. 2. Since angle is conguent to angle and since the vetical angles at ae equal, by, is simila to 2. Use atios geneated by these simila tiangles: / = / o * = *. 3. * = *. Let = x, then = + = + 7; 8 * 8 = * ( + 7). Solving: = 9. Since = 60 o, = 9, and = 8, is a tiangle with ight angle. Theefoe, = 9 3. nd is also a ight angle. So, = = 292 = Since is a ight angle, is a diamete of the cicle and it adius is 73. The aea of the cicle = π 2 = 73 π LMS. Let = x, then = 2x. * = * o 2x * x = 2 * 2; 2x 2 = 24; x = 2 3 ; = 3x = 2 3. Let W = x, then X = x + 2. Y*Z = W*X. 4*2 = x(x + 2); x 2 + 2x 48 = 0 = (x + 8)(x 6). W = 6. Let HK = x. 2 = H * K. 6 2 = 2(x + 2). X = 6 and K = 8. K = = 360 = Let x = = = ; y = =. * = *. x * 3x = y * 2y; 3x 2 = 2y 2 ; / = y/x = 3 2 o Let adius =, then = ; and = + ; * = * = ( )( + ) = simila solution yields: * = = * but Q 2 also equals * ; hence, = Q... See diagam. o a tangent and chod, the ltenate Segment Theoem states that fo any point on altenate side of cicle, = 2. Tiangles and by the theoem fo simila tiangles. 3. Use atios geneated by these simila tiangles.. y owe of oint, 2 = * and 2 = * ; hence, * = * 6. y ythagoean Theoem, =. * = *. 3* = * = 3 = 3 / = 3 2 = 3/2. ; = = / = 3 = 2

5 7. om 3, above, * = 2 2, 28 * 2 = = = 68; = 4 8. Since no dimensions ae given, WLG let = = =. Let Q = d, Q = d, = x, and = y. We must show that: = +. Via Simila tiangles, we will hunt fo /x and /y. Q x y ngles and ae conguent -- each subtends. With angle Q shaed, Q Q. / = Q/Q; /y = ( d)/q. Similaly, angles and ae conguent and Q Q. / = Q/Q; /x = d/q. Theefoe, + = + = d + d = x y Q Q Q 9. WLG, let = = = 2, then = = =. Let = x = G. Using owe of oint on : * = G * o * = (x+) * x. So: x 2 + x = 0. y quadatic fomula, x = (- + )/2. Q G d x Q d y x / = /x = = ( + )/2, the Golden atio. 0. HSM, 990, #20 Since and ae ight angles, they ae each on the cicle with diamete. Theefoe,,,, and ae cyclic and owe of oint can be applied. xtend to X on the cicle. Since X is on the cicle with diamete, X is a ight tiangle and X is a ight angle. Thus X is a ectangle with = X. pply owe of oint to. X 3 7 X * = * = X = 3 * 7/ = 2/.

4.3 Area of a Sector. Area of a Sector Section

4.3 Area of a Sector. Area of a Sector Section ea of a Secto Section 4. 9 4. ea of a Secto In geomety you leaned that the aea of a cicle of adius is π 2. We will now lean how to find the aea of a secto of a cicle. secto is the egion bounded by a cental